Assume $U$ and $H$ are Hilbert spaces and assume $U\subseteq H$ is continuous i.e. $U\subseteq H$ and $$\|x\|_H\leq c\|x\|_U\tag{$*$}$$
for some constant $c>0$ and all $x\in H$. If $H$ is seperable, then $U$ is seperable. So both spaces can be identified with $\ell^2$ for two unitary operators $T_1:H\rightarrow \ell^2$, $T_2: U\rightarrow \ell^2$.
I want to prove that $U$ is dense in $H$ in that case or are there counter examples that shows that this is wrong?.
Intuitively I think this could be true since:
If $x\in H$ then $T_1x \in \ell^2$ and there exists an approximating sequence $x_n$ in $\ell^2$ of $T_1x$, such that $T_2^{-1}x_n$ is an approximating sequence of $T_2^{-1}T_1x$ in $U$ and by $(*)$, $T_2^{-1}T_1x$ would be approximated by $T_2^{-1}x_n \in U$ in $H$.
No, you cannot deduct from a continuous embedding of $U$ into $H$ that $U$ is dense in $H$.
If $U$ is too meager, then $U$ cannot be dense in $H$.
E.g. Let us assume that $H$ is a separable infinite dimensional Hilbert space over $\mathbb{R}$ and pick some $u\in H$. Then set $U:=\{\lambda u : \lambda \in \mathbb{R}\}$ and equip it with the scalar product of $H$, thus $U$ is a (trivial) Hilbert space. Clearly $U$ satisfies the continuous embedding properties but is by far not dense in $H$.